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In a XY Plan, Y is above the line -x/2. I prefer to solve this DS with the XY plan.

Stat1 y > -1. This brings us to y above the line y=-1. The area covered could be either above -x/2 (when x>0) or below -x/2 (wehn x<10). (Fig1)

INSUFF.

Stat2 (2^x) - 4 >0
<=> 2^x > 4 = 2^2
<=> x > 2

In the XY plan, it does mean that the concerned area is at right of the line x=2. One more time, their is 2 case, y could be either above the line -x/2 (when y > 0) or below the line -x/2 (the point (3,-100)). (Fig2)

INSUFF.

Both (1) and (2): x > 2 and y > -1 represents an area with a not attainable vertice at (-1;2). By drawing the line -x/2 and this point, we observe that this point is on the line. Thus, the whole area of points such that x>2 and y>-1 is above the line -x/2. (Fig3)

Re: DS: Property of Number [#permalink]
03 Dec 2010, 10:22

Expert's post

devilmirror wrote:

Is -x < 2y (1) y > -1 (2) (2^x) - 4 > 0

Quite an old thread is resurrected.

Is -x<2y?

Is \(-x<2y\)? --> rearrange: is \(x+2y>0\)?

(1) y>-1, clearly insufficient as no info about \(x\).

(2) (2^x)-4>0 --> \(2^x>2^2\) --> \(x>2\) --> also insufficient as no info about \(y\).

(1)+(2) \(y>-1\), or \(2y>-2\) and \(x>2\) --> add this inequalities (remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(2y+x>-2+2\) --> \(x+2y>0\). Sufficient.

Re: DS: Property of Number [#permalink]
06 Feb 2014, 07:41

Bunuel wrote:

devilmirror wrote:

Is -x < 2y (1) y > -1 (2) (2^x) - 4 > 0

Quite an old thread is resurrected.

Is -x<2y?

Is \(-x<2y\)? --> rearrange: is \(x+2y>0\)?

(1) y>-1, clearly insufficient as no info about \(x\).

(2) (2^x)-4>0 --> \(2^x>2^2\) --> \(x>2\) --> also insufficient as no info about \(y\).

(1)+(2) \(y>-1\), or \(2y>-2\) and \(x>2\) --> add this inequalities (remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(2y+x>-2+2\) --> \(x+2y>0\). Sufficient.

Answer: C.

I chose C.. thats correct.. bt with different approach

bt Bunuel I didnt get this highlighted thing?? How did u do that? _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Re: DS: Property of Number [#permalink]
07 Feb 2014, 04:12

Expert's post

sanjoo wrote:

Bunuel wrote:

devilmirror wrote:

Is -x < 2y (1) y > -1 (2) (2^x) - 4 > 0

Quite an old thread is resurrected.

Is -x<2y?

Is \(-x<2y\)? --> rearrange: is \(x+2y>0\)?

(1) y>-1, clearly insufficient as no info about \(x\).

(2) (2^x)-4>0 --> \(2^x>2^2\) --> \(x>2\) --> also insufficient as no info about \(y\).

(1)+(2) \(y>-1\), or \(2y>-2\) and \(x>2\) --> add this inequalities (remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(2y+x>-2+2\) --> \(x+2y>0\). Sufficient.

Answer: C.

I chose C.. thats correct.. bt with different approach

bt Bunuel I didnt get this highlighted thing?? How did u do that?

\(2y+x>-2+2\) --> re-arrange the left hand side as x+2y. As for the right hand side: -2 + 2 = 0. So, we get \(x+2y>0\).

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